Question 107935
The diagonal of a rectangle is {{{2 ft}}} longer than its {{{length}}} and {{{5 ft}}} longer than its {{{width}}}. 
What are the dimensions of the rectangle?

Since the {{{diagonal}}} divides rectangle into two {{{right}}}{{{trianles}}}, we will use {{{Pythagoras' }}}{{{Theorem }}} in order to find the dimensions of the rectangle.

In algebraic terms, {{{a2 + b2 = c2}}}where {{{c}}} is the hypotenuse ( in your case the {{{diagonal}}}) while {{{a}}} and {{{b }}}are the legs of the triangle (in your case the {{{lenght}}} and {{{width}}}).

given:

{{{c = a + 2ft}}}   => {{{a = c – 2ft}}}

{{{c = b + 5ft}}}  =>  {{{b = c – 5ft}}}


we will substitute it in:

{{{c^2 = a^2 + b^2}}}

{{{c^2 = (c-2ft)^2 + (c-5ft)^2}}}

{{{c^2 = c^2 -4cft + 4ft^2 + c^2 – 10cft + 25ft^2}}}

{{{c^2 = 2c^2 – 14cft +29ft^2}}}

{{{c^2 -14cft + 29ft^2 = 0}}}

{{{c[1,2]=(14 +- sqrt (14^2 -4*1*29 )) / (2*1)}}}

We need only positive root

{{{c[1]=(14 + sqrt (14^2 -4*1*29 )) / (2*1)}}}


{{{c[1]=(14 + sqrt (196-116  )) / 2}}}

{{{c[1]=(14 + sqrt (80)) / 2}}}


{{{c[1]=(14 + 8.9) / 2}}}

{{{c[1]=(22.9) / 2}}}

{{{c[1]=(11.45ft) }}}



{{{a = c – 2ft}}}
=> {{{a = 11.45ft – 2ft = 9.45ft}}}

{{{b = c – 5ft}}}
{{{b = 11.45ft – 5ft}}}
=> {{{b = 6.45ft}}}

Check:

{{{(11.45)^2 = (9.45)^2 + (6.45)^2}}}

{{{131.1025 = 89.3025 + 41.6025}}}

  {{{131 = 131}}}